What are the trigonometry questions for arbitrary angles

Classic trigonometric functions

Classic trigonometric functions are as follows:

1, the basic relationship between trigonometric functions of the same angle

Inverse relationship:tanα-cotα=1sinα-cscα=1cosα-secα=1; quotient relationship:sinα/cosα=tanα=secα/cscα cosα/sinα = cotα = cscα/secα square relationship: sin^2(α) + cos^2(α) = 11 + tan^2(α) = sec^2(α)1 + cot^2(α) = csc^2(α)

2, the usual two formulas for the different conditions of common use: sin²α + cos²α =1tanα*cotα=1

3, a special formula: (sina+sinθ)*(sina+sinθ)=sin(a+θ)*sin(a-θ)

Proof: (sina+sinθ)*(sina+sinθ)=2sin[(θ+a)/2]cos[(a-θ)/2]* 2cos[(θ+a)/2]sin[(a-θ)/2]=sin(a+θ)*sin(a-θ)

4, acute trigonometric function formula. Sine: sinα = opposite side of ∠α / hypotenuse of ∠α; cosine: cosα = neighbor of ∠α / hypotenuse of ∠α; tangent: tanα = opposite side of ∠α / neighbor of ∠α; cotangent: cotα = neighbor of ∠α / opposite side of ∠α.

There are a few high school math problems I’d like to ask you about (trigonometry on arbitrary angles)

1And one more question how come a isn’t used later in the first question?


Obviously B

3. sin x-cos x-tan x=(sinx)², so as long as sinx is not equal to 0 it’s fine, so go with B


(sin 330°-tan ((13/-3)π)/cos ((19/-6)π )-cos 690°)

(sin -30°-tan ((2/3)π)/cos ((5/6)π)-cos -30°)

(-1/2 – – √(3)) / – √(3)/2 – 1/2)


4. tan 136°=tan-44<0 sin 288°=sin-72<0 cos 188°=-cos8<0, so multiply <0

5. sin @-cos @ <0, so sincos isosign, for two or four quadrants, and if in four quadrants, cos is positive and sin is negative

Positive minus negative can’t be less than zero, so it’s in the second quadrant B

What is the trigonometric function of an arbitrary angle?

The value of the trigonometric function of an arbitrary angle can be found one or more times by using the trigonometric function of the sum (or difference) of the special angle, and the half angle of the special angle.

For example:






Same-angle trigonometric functions

(1) Square relationship:




(2) Relationships of products:




High school math trigonometry arbitrary angle questions with high scores!

First find A∩B,change the letter,replace k in B with m.

From k*135=m*150,m=-10,-9,-8,… ,8

You can get 9k=10m, and 9 and 10 are mutually prime, so m can only take a multiple of 9 for the equation to hold.

That is, m=-9,0 two numbers, and the corresponding k=-10,0.

So A∩B={-1350°, 0°}

So the answer sought is:

S={x|x=-1350°+n*360°, n is an integer}∪{x|x=n*360°, n is an integer}

Or Write S={x|x=-90°+n*360°,n is an integer}∪{x|x=n*360°,n is an integer}